Question

find the product.
(4p - 3) + q²
(4p - 3) + q² = \square

Explanation:

Step1: Apply square of sum formula

Recall $(a+b)^2=a^2+2ab+b^2$, let $a=4p-3$, $b=q$.
$$[(4p-3)+q]^2=(4p-3)^2+2(4p-3)q+q^2$$

Step2: Expand $(4p-3)^2$

Use $(c-d)^2=c^2-2cd+d^2$, $c=4p$, $d=3$.
$$(4p-3)^2=(4p)^2-2(4p)(3)+3^2=16p^2-24p+9$$

Step3: Expand $2(4p-3)q$

Distribute the 2q to terms inside.
$$2(4p-3)q=8pq-6q$$

Step4: Combine all expanded terms

Add the results from Steps 2 and 3 with $q^2$.
$$16p^2-24p+9+8pq-6q+q^2$$

Step5: Rearrange terms (standard order)

Group like terms and arrange by degree.
$$16p^2+8pq+q^2-24p-6q+9$$

Answer:

$16p^2 + 8pq + q^2 - 24p - 6q + 9$